Reading the series as
[tex]\displaystyle\sum_{n\ge1}\frac{n!}{n^n}[/tex]
With the summand [tex]a_n=\dfrac{n!}{n^n}[/tex], we have
[tex]\left|\dfrac{a_{n+1}}{a_n}\right|=\left|\dfrac{\frac{(n+1)!}{(n+1)^{n+1}}}{\frac{n!}{n^n}}\right|=(n+1)\dfrac{n^n}{(n+1)^{n+1}}=\dfrac{n^n}{(n+1)^n}=\left(\dfrac n{n+1}\right)^n[/tex]
Then
[tex]\displaystyle\lim_{n\to\infty}\left(\frac n{n+1}\right)^n=\lim_{n\to\infty}e^{n\ln\frac n{n+1}}=\exp\left(\lim_{n\to\infty}n\ln\frac n{n+1}\right)[/tex]
This is a pretty standard limit that utilizes L'Hopital's rule. We can write
[tex]\displaystyle\lim_{n\to\infty}n\ln\frac n{n+1}=\lim_{n\to\infty}\frac{\ln\frac n{n+1}}{\frac1n}\stackrel{\mathrm{LHR}}=\lim_{n\to\infty}\frac{\frac1{n(n+1)}}{-\frac1{n^2}}=-1[/tex]
which means
[tex]\displaystyle\lim_{n\to\infty}\left(\frac n{n+1}\right)^n=e^{-1}<1[/tex]
and so the series converges by the ratio test.
